Poj 2540 hotter colder cut Polygon

/* Poj 2540 hotter colder cut a polygon using a vertical line of two points to cut the polygon. Based on the cold and hot data, determine the half and then output the area */# include <stdio. h> # include <math. h> const double EPS = 1e-8; const int n = 200; struct point {Double X, Y; point () {} Point (double A, double B ): X (A), y (B) {}} Dian [N]; point Jiao [N]; inline bool mo_ee (Double X, Double Y) {double ret = x-y; If (Ret <0) ret =-ret; If (Ret <EPS) return 1; return 0;} inline bool mo_gg (Double X, double Y) {return x> Y + EPS;} // X> Y inline bool mo_ll (Double X, Double Y) {return x <Y-EPS;} // x <y inline bool mo_ge (Double X, Double Y) {return x> Y-EPS;} // x> = y inline bool mo_le (Double X, Double Y) {return x <Y + EPS ;} // x <= y inline double mo_xmult (point P2, point P0, Point P1) // P1 returns a negative value on the left of P2 and a positive {return (p1.x-forwarded X) on the Right) * (p2.y-p0.y)-(p2.x-p0.x) * (p1.y-p0.y);} Point mo_intersection (point u1, point U2, point V1, Po Int V2) {point ret = U1; Double T = (u1.x-v1.x) * (v1.y-v2.y)-(u1.y-v1.y) * (v1.x-v2.x )) /(u1.x-u2.x) * (v1.y-v2.y)-(u1.y-u2.y) * (v1.x-v2.x); ret. X + = (u2.x-u1.x) * t; ret. Y + = (u2.y-u1.y) * t; return ret ;} ////////////// // method vector point mo_getfaxian (point xiang) {point; if (mo_ee (Xiang. x, 0) {. X = 1;. y = 0; return a;} else if (mo_ee (Xiang. y, 0) {. X = 0;. y = 1; return a;} else {. X = 1;. y =-1.0 * Xiang. x/x IANG. y; return a ;}/// returns the polygon area. Double mo_area_polygon (point * Dian, int N) {int I; point yuan; yuan. X = yuan. y = 0; double ret = 0; for (I = 0; I <n; ++ I) {RET + = mo_xmult (Dian [(I + 1) % N], yuan, Dian [I]);} If (Ret <0) ret =-ret; return ret/2;} Point mo_banjiao_jiao_temp [N * 2]; void mo_banjiao_cut (point * ans, point Qian, point Hou, Int & nofdian) {int I, K; for (I = k = 0; I <nofdian; ++ I) {double A, B; A = mo_xmult (Hou, ANS [I], Qian); B = mo_xmult (Hou, ANS [(I + 1) % nofdian], Qian); If (mo_ge (A, 0 )) // clockwise <= 0 {mo_banjiao_jiao_temp [k ++] = ans [I];} If (mo_ll (A * B, 0 )) {mo_banjiao_jiao_temp [k ++] = mo_intersection (Qian, Hou, ANS [I], ANS [(I + 1) % nofdian]) ;}} for (I = 0; I <K; ++ I) {ans [I] = mo_banjiao_jiao_temp [I];} nofdian = K;} int main () {point Qian (0, 0); point cur, mid, end; char order [20]; int flag = 0; Jiao [0] = point (0, 0); Jiao [1] = point (10, 0 ); jiao [2] = point (10, 1 0); Jiao [3] = point (0, 10); int jiaodian = 4; while (scanf ("% lf", & cur. X )! = EOF) {scanf ("% lf % s", & cur. y, order); getchar (); If (Order [0] = 's' | flag = 1) {flag = 1; printf ("0.00 \ n"); continue;} mid. X = (cur. X + Qian. x)/2; mid. y = (cur. Y + Qian. y)/2; end = mo_getfaxian (point (cur. x-qian.x, cur. y-qian.y); end. X = mid. X + end. x; end. y = mid. Y + end. y; bool Zai = mo_gg (mo_xmult (end, cur, mid), 0); If (Order [0] = 'H' & Zai) | (Order [0] = 'C '&&(! Zai) {} else {point TEM = end; end = mid; Mid = TEM;} mo_banjiao_cut (Jiao, mid, end, jiaodian); double area = mo_area_polygon (Jiao, jiaodian); printf ("%. 2lf \ n ", area); Qian = cur;} return 0 ;}